Mathematics Advanced • Year 11 • Module 3 • Lesson 2

Limits

Apply limit evaluation to multi-step problems involving secant gradients, removable discontinuities, piecewise pricing models and limit laws.

Apply · Problem Set

Problem 1, Bridging a hole in the graph

Consider the function

f(x) = (x² − 4) / (x − 2), x ≠ 2.

Set up: What are we solving for?

(i) Show that direct substitution gives 0/0, and use factor-and-cancel to evaluate lim_x→2 f(x). 2 marks

(ii) Sketch (in the space) the graph of f near x = 2, marking the hole at x = 2 with an open circle and the limit value with a small horizontal arrow. 2 marks

(iii) Define a new function g(x) that agrees with f for x ≠ 2 but is continuous everywhere. State the single extra rule needed and justify in one sentence. 2 marks

Stuck? Revisit lesson § Removable discontinuity.

Problem 2, A piecewise pricing model

An electricity provider charges a unit price per kWh that changes at usage x = 200 kWh per month:

P(x) = { 0.32 x if x ≤ 200 ; 64 + 0.40(x − 200) if x > 200 }

(P in dollars per month, x in kWh.)

Set up: What are we solving for?

(i) Compute lim_x→200⁻ P(x) and lim_x→200⁺ P(x). 2 marks

(ii) Hence state whether lim_x→200 P(x) exists, and whether P is continuous at x = 200. 2 marks

(iii) The company wants the bill to be continuous at x = 200 (no sudden jump). They keep the 0.32 rate below 200 kWh but propose changing the upper-bracket constant. What value should replace the 64 in the upper formula to guarantee continuity, and why? 2 marks

Problem 3, A secant gradient as a limit

For f(x) = x², the gradient of the secant joining the points (3, 9) and (3 + h, f(3 + h)) where h ≠ 0 can be written

m(h) = ( f(3 + h) − f(3) ) / h.

Set up: What are we solving for?

(i) Show that m(h) simplifies to 6 + h (for h ≠ 0), showing every line of algebra. 2 marks

(ii) Hence evaluate lim_h→0 m(h). State what this limit represents geometrically. 2 marks

(iii) Tabulate m(h) for h = 0.1, 0.01, 0.001 and confirm the numerical values match your algebraic answer to (ii). 2 marks

Stuck? You did the same algebra in Lesson 1 Worksheet 2, but now framed as a limit.

Problem 4, Limit laws and a quotient with a hole

You are given lim_x→2 f(x) = 3 and lim_x→2 g(x) = 5.

Set up: What are we solving for?

(i) Using only the limit laws (no specific f, g formula), evaluate each of:

(a) lim_x→2 [ f(x) + 2 g(x) ] = ____________ (b) lim_x→2 [ f(x) · g(x) ] = ____________

(ii) Why does the quotient law fail to apply to lim_x→3 [ (x − 3) / (x² − 9) ] by direct substitution? Show how to rearrange before applying limit laws, then evaluate. 3 marks

(iii) A student writes lim_x→0 [ x · (1/x) ] = lim_x→0 x · lim_x→0 (1/x) and concludes the limit is "0 × ∞ = 0". Explain in one sentence why this is wrong and state the correct value. 2 marks

Problem 5, Choose the constant to fit the function

Consider the piecewise function

f(x) = { x² + k if x ≤ 1 ; 3x if x > 1 }.

Set up: What are we solving for?

(i) Find lim_x→1⁻ f(x) and lim_x→1⁺ f(x) in terms of k. 2 marks

(ii) Hence find the value of k for which lim_x→1 f(x) exists, and state the value of the limit for that k. 2 marks

(iii) Using that value of k, is f continuous at x = 1? Justify by checking both: (a) the limit exists and (b) the limit equals f(1). 2 marks

Stuck on (iii)? f(1) uses the top branch (≤ 1). Compare against the limit you found in (ii).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

Problem 1, Removable hole at x = 2

Set up. We confirm that f has a removable discontinuity at x = 2, find the natural value of the limit there, and patch the hole to make a continuous extension.

(i) Substituting x = 2: (4 − 4)/(2 − 2) = 0/0. Factor: x² − 4 = (x − 2)(x + 2); cancel (x − 2) (for x ≠ 2): expression = x + 2. lim_x→2 (x + 2) = 4.

(ii) The graph is the line y = x + 2 everywhere except an open circle at (2, 4). Sketch should show a straight line of gradient 1 passing through (0, 2), with a small open circle at (2, 4) and arrows from both sides converging to that y-value.

(iii) Define g(x) = f(x) for x ≠ 2 and g(2) = 4. Justification: this single redefinition is exactly the limit value, so the left limit, right limit and function value all agree, continuity restored.

Problem 2, Electricity tariff piecewise

Set up. Check whether the two branches of the tariff meet at the boundary x = 200. If they do, the limit exists and the bill is continuous; if not, the customer sees a jump in the bill at the bracket boundary.

(i) Left limit: lim_x→200⁻ 0.32x = 0.32(200) = $64. Right limit: lim_x→200⁺ [64 + 0.40(x − 200)] = 64 + 0.40(0) = $64.

(ii) Left = right = 64, so lim_x→200 P(x) exists and equals $64. Since P(200) = 0.32(200) = 64 (top branch, x ≤ 200), the limit equals the function value, so P is continuous at x = 200.

(iii) The 64 in the upper formula was chosen specifically to equal 0.32 × 200, the value of the lower branch at x = 200. Any other constant would break the matching at the boundary and produce a jump discontinuity in the bill (the customer would see a sudden jump in their monthly charge). So the constant must remain 64; any change would cause a discontinuity.

Problem 3, Secant gradient of x² at x = 3

Set up. Compute the gradient of the secant joining (3, 9) and (3 + h, f(3 + h)) as a function of h, then take the limit as h shrinks.

(i) f(3 + h) = (3 + h)² = 9 + 6h + h². m(h) = ((9 + 6h + h²) − 9) / h = (6h + h²) / h = h(6 + h) / h = 6 + h (h ≠ 0).

(ii) lim_h→0 (6 + h) = 6. Geometrically: this is the gradient of the tangent to y = x² at the point (3, 9), i.e., the instantaneous rate of change of x² at x = 3.

(iii) m(0.1) = 6.1; m(0.01) = 6.01; m(0.001) = 6.001, all clearly approaching 6, confirming the algebraic answer in (ii).

Problem 4, Limit laws

Set up. Apply each law term-by-term, then identify where direct substitution fails.

(i) (a) 3 + 2(5) = 13. (b) 3 × 5 = 15. (c) 5 / 3 = 5/3. (d) 3³ = 27.

(ii) Direct substitution: (3 − 3) / (9 − 9) = 0/0. The quotient law requires lim of denominator ≠ 0, so it does not apply directly. Rearrange: (x − 3) / (x² − 9) = (x − 3) / ((x − 3)(x + 3)) = 1 / (x + 3) (for x ≠ 3). Now lim_x→3 1 / (x + 3) = 1 / 6 (quotient law applies, denominator is 6 ≠ 0). Answer: 1/6.

(iii) The product law lim[f · g] = lim f · lim g requires both individual limits to exist. lim_x→0 (1/x) does not exist (it heads to +∞ on one side, −∞ on the other), so the law does not apply. The actual value of x · (1/x) is 1 for every x ≠ 0, so lim_x→0 [x · (1/x)] = lim_x→0 1 = 1.

Problem 5, Choose k for the piecewise function

Set up. Match the two one-sided limits at x = 1, then choose k to make them agree (existence). Finally compare against f(1) to check continuity.

(i) Left: lim_x→1⁻ (x² + k) = 1 + k. Right: lim_x→1⁺ 3x = 3.

(ii) Limit exists ⇔ 1 + k = 3, so k = 2. With k = 2, lim_x→1 f(x) = 3.

(iii) With k = 2, f(1) = 1² + 2 = 3 (top branch). Both conditions: (a) limit exists and equals 3 ✓; (b) limit equals f(1) = 3 ✓. Hence f is continuous at x = 1.